Question

Verify that the most probable speed of an ideal gas molecule is √ 2kT m .

Answer 1

We have to verify that the most probable speed of an ideal gas molecule is $\sqrt{\frac{2kT}{m}}$.

we know according to Maxwell - Boltzmann's distribution,

$\frac{dN}{N}=\left(\frac{m}{2\pi k_BT}\right)^{1/2}e^{\frac{-mv^2}{2k_BT}}dv$.....(1)

where dN/N is the fraction of molecules moving at velocity v to v + dv.

m is the mass of molecules

$k_B$ is Boltzmann's constant

and T is absolute Temperature.

equation (1) can be written in terms of the scalar quantity instead of the vector quantity. The function is given by,

$f(c)=4\pi c^2\left(\frac{m}{2\pi k_BT}\right)^{3/2}e^{\frac{-mc^2}{2k_BT}}$

This form of function defines the distribution of the gas molecules moving at different speeds.

Now find the most probable speed :

we know, the most probable speed is the maximum speed attained by maximum number of gaseous molecule. this is established by finding the speed when the following derivatives is zero.

i.e., $\left|\frac{df(c)}{dc}\right|_{C_m}=0$

we get, $C_m=\sqrt{\frac{2k_BT}{m}}$

Therefore the most probable speed of an ideal gas molecule is $\sqrt{\frac{2kT}{m}}$